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Double Integrals on Rectangles · Tutorial 505 of 999

Signed Volume Under a Surface Over a Rectangle

Learn to read a double integral over a rectangle as net signed volume, distinguishing the contributions above and below the \(xy\)-plane.

Beginner 9 min read

What You'll Learn

  • Interpret the height of a surface as a signed contribution to volume.
  • Distinguish signed volume from ordinary, nonnegative geometric volume.
  • Determine how regions above and below the \(xy\)-plane contribute to a double integral.
  • Use Riemann sums to justify the signed-volume interpretation.
  • Recognize when positive and negative volumes cancel.
  • Track the units of a double integral as area times height.

From Surface Height to Signed Volume

A double integral over a rectangle combines the values of a function across a two-dimensional region. One way to understand that accumulation is to picture the graph \(z=f(x,y)\) above the rectangle in the \(xy\)-plane.

At each point \((x,y)\), the function value \(f(x,y)\) is the vertical height of the graph above the plane \(z=0\). A positive height lies above the plane; a negative height lies below it. As a result, the integral records a signed volume: contributions above the plane count positively, while contributions below it count negatively.

Key idea: The double integral measures net signed volume, not necessarily the total geometric volume between the surface and the \(xy\)-plane.

The Signed-Volume Interpretation

Let \(R=[a,b]\times[c,d]\) be a rectangle, and let \(f\) be integrable on \(R\). Partition \(R\) into smaller rectangles. For a small rectangle with area \(\Delta A\), choose a sample point \((x^*,y^*)\). The graph’s height there is \(f(x^*,y^*)\), so the corresponding signed box has contribution \(f(x^*,y^*)\Delta A\).

If that height is positive, the box extends above the \(xy\)-plane and its contribution is positive. If the height is negative, the box extends below the plane and its contribution is negative. Adding these contributions gives a Riemann sum. As the partition is refined, the sums approach the double integral.

$$ \iint_R f(x,y)\,dA = \lim_{\text{mesh}\to 0} \sum_{i,j} f(x_{ij}^*,y_{ij}^*)\,\Delta A_{ij}. $$

Here, \(\Delta A_{ij}\) is the area of the \(ij\)-th small rectangle, and the mesh is the largest diameter of any rectangle in the partition. This limit is the definition of the double integral as a limit of Riemann sums. The signed-volume interpretation follows because each term in the sum is the signed volume of one approximating box.

For a surface that lies entirely above the plane, the integral is nonnegative and equals the volume between the graph and the plane. If the surface lies entirely below the plane, the integral is nonpositive: its magnitude is the geometric volume below the plane, but its sign is negative. If the surface crosses the plane, the positive and negative contributions combine to give a net result.

Definition: For an integrable function \(f\) on \(R\), the signed volume under \(z=f(x,y)\) over \(R\) is \(\iint_R f(x,y)\,dA\). Geometric volume is nonnegative; signed volume counts the part above \(z=0\) positively and the part below \(z=0\) negatively.

Why the Sign Appears

The sign is already present in the height, before any volumes are added. A box with base area \(\Delta A\) and height \(h\) contributes \(h\Delta A\), not \(|h|\Delta A\). Replacing \(h\) by its absolute value would turn every box into a positive contribution and would compute a different quantity.

To see the reasoning precisely, consider one partition and its Riemann sum. Separate its terms according to whether the sampled height is positive or negative. The sum is the total of the positive box contributions minus the magnitudes of the negative box contributions. Taking the limit preserves this net-volume interpretation: that limit is the integral. Thus the interpretation is not an extra formula imposed on the integral; it is what the signed Riemann sums represent.

1
Read the height.
At each sample point, \(f(x,y)\) gives the signed vertical height above the \(xy\)-plane.
2
Form a signed box.
Multiply that height by the small rectangle’s area. A negative height gives a negative contribution.
3
Add and take the limit.
The signed box contributions form Riemann sums; their limit is the signed volume.

Worked Example: A Constant Surface Below the Plane

Worked Example: A Constant Surface Below the Plane

Let \(f(x,y)=-2\) on \(R=[1,4]\times[-1,1]\). Find the signed volume and the geometric volume between the graph and the plane.

The side lengths of the rectangle are \(4-1=3\) and \(1-(-1)=2\), so its area is

$$ A(R)=3\cdot 2=6. $$

The surface has constant height \(-2\). Every box in every Riemann sum therefore has signed volume equal to its base area times \(-2\). The total signed volume is

$$ \iint_R -2\,dA=(-2)(6)=-12. $$

The graph is two units below the plane throughout the rectangle, so the ordinary geometric volume is \(2\cdot6=12\) cubic units. The signed-volume answer is \(-12\) cubic units. As a check, its magnitude is \(12\), matching the geometric volume, and its negative sign correctly records that the surface lies below the plane.

Worked Example: Equal Volumes Can Cancel

Worked Example: Equal Volumes Can Cancel

Let \(f(x,y)=x-1\) on \(R=[0,2]\times[0,3]\). Find the signed volume and the total geometric volume between the graph and the plane.

The height is zero at \(x=1\), negative when \(0\le x<1\), and positive when \(1<x\le2\). The rectangle therefore divides into two parts of equal base dimensions: each part has width \(1\) in the \(x\)-direction and length \(3\) in the \(y\)-direction.

On the positive side, the height rises from \(0\) to \(1\) as \(x\) goes from \(1\) to \(2\). In an \(xz\)-cross-section, the region under the graph is a triangle with base \(1\) and height \(1\). Its area is

$$ \frac12\cdot1\cdot1=\frac12. $$

This cross-section extends a distance \(3\) in the \(y\)-direction, so the volume above the plane is \(\frac12\cdot3=\frac32\) cubic units.

On the negative side, the magnitude of the height rises from \(0\) to \(1\) as \(x\) goes from \(1\) to \(0\). Its triangular cross-section also has area \(\frac12\cdot1\cdot1=\frac12\), and its length in the \(y\)-direction is \(3\). The volume below the plane is therefore \(\frac12\cdot3=\frac32\) cubic units.

The signed volume is the positive contribution minus the magnitude of the negative contribution:

$$ \iint_R (x-1)\,dA = \frac32-\frac32 = 0. $$

The total geometric volume, which counts both parts positively, is \(\frac32+\frac32=3\) cubic units. As a check, the equal positive and negative volumes cancel in the signed result, while their sum is the nonzero geometric volume.

Worked Example: A Surface Entirely Above the Plane

Worked Example: A Surface Entirely Above the Plane

Let \(f(x,y)=x+y\) on \(R=[0,2]\times[0,1]\). Use midpoint Riemann sums to find the signed volume.

The function is nonnegative throughout the rectangle, so all the box contributions are nonnegative. Divide the \(x\)-interval into \(m\) equal parts and the \(y\)-interval into \(n\) equal parts. Each small rectangle has area

$$ \Delta A=\frac{2}{m}\cdot\frac{1}{n}=\frac{2}{mn}. $$

The midpoint coordinates are \(x_i=\frac{2}{m}(i-\frac12)\) for \(i=1,\ldots,m\), and \(y_j=\frac{1}{n}(j-\frac12)\) for \(j=1,\ldots,n\). Their sums are

$$ \sum_{i=1}^m x_i = \frac{2}{m}\left(\frac{m(m+1)}{2}-\frac{m}{2}\right) =m, \qquad \sum_{j=1}^n y_j = \frac{1}{n}\left(\frac{n(n+1)}{2}-\frac{n}{2}\right) =\frac{n}{2}. $$

The midpoint Riemann sum is

$$ \begin{aligned} S_{mn} &=\sum_{i=1}^{m}\sum_{j=1}^{n}(x_i+y_j)\frac{2}{mn}\\ &=\frac{2}{mn}\left(n\sum_{i=1}^{m}x_i+m\sum_{j=1}^{n}y_j\right)\\ &=\frac{2}{mn}\left(nm+\frac{mn}{2}\right)=3. \end{aligned} $$

Every such midpoint sum is \(3\), so its limit is \(3\). Therefore the signed volume is \(3\) cubic units. As a check, the result is positive, as expected because the surface is above the plane everywhere; the Riemann sums also give the value directly.

Signed Volume and Geometric Volume Are Different

For a surface that crosses the plane, the distinction matters. Signed volume measures the difference between the volume above the plane and the volume below it. Total geometric volume measures the sum of those two nonnegative volumes. In the second example, the signed result is zero even though the region between the surface and the plane has geometric volume \(3\).

A zero integral does not, by itself, mean that the graph lies in the plane or that there is no volume. It can instead mean that positive and negative contributions cancel. Likewise, a negative integral does not represent a negative physical volume: it indicates that the below-plane contribution is larger than the above-plane contribution.

Common Pitfall: Dropping the Sign

A common error is to report only a nonnegative volume when asked for a double integral. For the constant surface in the first example, one might calculate

$$ \text{wrong:}\qquad (-2)(6)=12. $$

That calculation has silently replaced the height \(-2\) by its magnitude \(2\). The correction is to retain the signed height:

$$ \text{correct:}\qquad (-2)(6)=-12. $$

The geometric volume is \(12\), but the signed volume, and hence the double integral, is \(-12\). Before interpreting an answer, check which quantity is requested and whether the graph is above or below the \(xy\)-plane.

Units and Interpretation

Each Riemann-sum contribution is height times base area. If both horizontal coordinates are measured in metres and the height is also measured in metres, then \(\Delta A\) has units of square metres and \(f(x,y)\Delta A\) has units of cubic metres. More generally, the units of the integral are the units of height multiplied by the units of area.

This is useful whenever a function describes a varying depth, elevation, or other signed height over a planar region. The integral accumulates that height across the rectangle. Its sign tells whether the net contribution lies above or below the reference plane; its magnitude alone does not tell whether positive and negative regions have cancelled.

A Practical Reading Checklist

  • Identify the rectangle over which the function is defined.
  • Check where \(f(x,y)\) is positive, negative, or zero.
  • Interpret positive heights as contributions above the plane and negative heights as contributions below it.
  • Use the integral as net signed volume; do not replace signed heights by their absolute values.
  • If total geometric volume is wanted, account for both sides of the plane as nonnegative volumes.
  • Check that the units are height times area.

Check Your Understanding

Use the sign of the function and the signed-volume interpretation to answer these questions.

  1. If \(f\) is negative everywhere on a rectangle, what is the sign of its double integral, and how does its magnitude relate to geometric volume?
  2. A surface has geometric volume \(5\) above the plane and \(2\) below it. What is its net signed volume?
  3. Can a double integral be zero when the surface is not the \(xy\)-plane? Explain how.
  4. For a constant height of \(3\) over a rectangle of area \(4\), what is the signed volume?
  5. Why does replacing every height by its absolute value generally change the quantity represented by the double integral?

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